JEE Main 2020 — Electrostatics Question with Solution

Let particle have charge q and mass 'm

We know, F = ma_y = qE

$$\therefore$$ $$a_y=\frac{qE}{m}$$

As electric field is along vertical direction so horizontal component of acceleration is zero.

$$\therefore$$ a_x = 0

Applying, $$S=u_{y}t+\frac{1}{2}{a}_y{t}^2$$

$$-y=0\times t+\frac{1}{2}\left(\frac{qE}{m}\right)t^2$$ ....(1)

Also, $$x=V_{0}t$$

$$\Rightarrow t=\frac{x}{V_0}$$

Putting in (1) we get,

$$y=-\frac{1}{2}\left(\frac{qE}{m}\right){\left(\frac{x}{V_0}\right)}^2$$

Differentiating with respect to x,

$$\frac{dy}{dx}=-\frac{qE}{m}\left(\frac{x}{V_0^2}\right)$$

$$\therefore$$ Slope = $$m=\tan \theta =-\frac{qE}{m}\left(\frac{x}{V_0^2}\right)$$

when x = d, then $$m=-\frac{qE}{m}\left(\frac{d}{V_0^2}\right)$$

We know, equation of straight line whose slope is m,

y = mx + c

Here m = $$\left(-\frac{qEd}{m{V}_0^2}\right)$$,

$$\therefore$$ $$-\frac{1}{2}\left(\frac{qE}{m}\right){\left(\frac{d}{V_0}\right)}^2=\left(-\frac{qEd}{m{V}_0^2}\right)d+c$$

$$\Rightarrow$$ c = $$\frac{qEd}{2m{V}_0^2}$$

Putting the value of c, we get

$$y=-\left(\frac{qEd}{m{V}_0^2}\right)x+\frac{qE{d}^2}{2m{V}_0^2}$$

$$=\frac{qEd}{m{V}_0^2}\left(\frac{d}{2}-x\right)$$